Strength of materials - Torsion

hafts: the torsional moment or couple 6.1. Introduction Torsion members – the slender members subjected to torsional loading, that is loaded by couple that produce twisting of the member about its axis Examples – A torsional moment (torque) is applied to the lug-wrench shaft, the shaft transmits the torque to the generator, the drive shaft of an automobile... y x z F A B C Q1 Q2 1 T 1 t 2 T t 2 1/10/2013 7 6.1. Introduction  Internal torsional moment diagram • Using method of section zM > 0 y z x • Sign convention of Mz - Positive: clockwise - Negative: counterclockwise 0zM  Mz = z y x 1/10/2013 8 6.2.1. Simplifying assumptions 6.2. Torsion of Circular Shafts 1/10/2013 9 6.2. Torsion of Circular Shafts • Consider the portion of the shaft shown in the figure • CD – before deformation • CD’ – after deformation - From the geometry 'DD d dz    => The Shear strain: d dz    d G G dz       - Following Hooke’s law: 6.2.2. Compatibility - d – the angle of twist => In the cross-section, only shear stress exists 1/10/2013 10 2 z p A A d d M dA G dA G I dz dz          z p Md dz GI     z p M I    – radial position Ip – polar moment of inertia Mz – internal torsional moment 6.2. Torsion of Circular Shafts 6.2.3. Equilibrium 6.2.3. Torsion formulas – Shearing stress  – the rate of twist 1/10/2013 11 6.2. Torsion of Circular Shafts max z z p p M M R I W    - Maximum shearing stress – Angle of twist From 6.2.3:   0 A L z z AB p pB M dz M dz rad GI GI     L   A B O a b c - Wp - Section modulus of torsion 1/10/2013 12 z AB p M L GI  z p M const GI  6.2. Torsion of Circular Shafts If the shaft is subjected to several different torques or cross-sectional area, or shear modulus changes abruptly from one region of the shaft to the next. – Multiply torques constz p i M GI        1 n z AB i i p i M l GI            GIp – stiffness of torsional shaft 1/10/2013 13 6.2. Torsion of Circular Shafts 1/10/2013 14 6.3. Strength Condition – Stiffness condition  ax pW z m M      0 n    - Determine experimentally 0     2    - Third strength hypothesis 3     3    - Fourth strength hypothesis 4  Strength condition:  Stiffness condition:    ax ax /zm p m M rad m GI           1/10/2013 15 6.3. Strength Condition – Stiffness condition  Three main problems:  max 30.2D zM  For a circular shaft:  ax 4 0.1 z m M GD    1. Investigating the strength condition, (stiffness condition)  max 3 ???0.2D zM   2. Determine the diameter of circular cross-section   3D 0.2 zM   3. Determine the maximum torque   3M 0.2z D 1/10/2013 16 • Assume that the reactions at the fixed ends MA, MD are shown in the figure. • Equilibrium: MA + MD = M (1) • Compatibility condition: AD = 0 (2) d a 2a D M MA D A M B2 d D M DM z z CD 2AB BDz z AD AB BD AB BD p p M a M a GI GI       BD z DM M AB z DM M M      4 4 2 0 0,10,1 2 D D AD M M a M a G dG d         1 32 ; 33 33 D AM M M M  Mz M/33 32M/33 6.4. Statically Indeterminate Problem 1/10/2013 17 J T xy      dV GJ T dV G U xy 2 222 22  • For a shaft subjected to a torsional load, • Setting dV = dA dx,               L L A L A dx GJ T dxdA GJ T dxdA GJ T U 0 2 0 2 2 2 0 2 22 2 22   • In the case of a uniform shaft, GJ LT U 2 2  6.5. Strain Energy 1/10/2013 18 6.6. Example Problem 1 • A Circular shaft made from two segments, each having diameter of D and 2D. The Shaft is subjected to the torques shown in the figure. 1. Draw the internal torsional moment diagram 2. Determine the maximum shearing stress 3. Determine the angle of twist of the end D with respect to B With M=5kNm; a=1m; D=10cm; G=8.103 kN/cm2 2a B a C D D M 3M 2 D 1/10/2013 19 1. Internal torsional moment diagram Segment CD Segment BC 2a B a C D D M 3M 2 D Mz kNm 15 10  10  z a 3 15CDzM M kNm  2 10BCzM M kNm   20 2 z a Mz CD 3M z1 3MM z2 a Mz BC 6.6. Example Problem 1 1/10/2013 20 6.6. Example Problem 1 2. Maximum shearing stress 3. Angle of twist of end D 2a B a C D D M 3M 2 D Mz kNm 15 10 2 max 2 3 3 15 10 7,5( / ) 0,2 0,2 10CD CD zM kN cm D         2 max 2 3 3 10 10 0,625( / ) 0,2 200,2 2 BC BC zM kN cm D       D BC CD    2CD BCz z D CD BC p p M a M a GI GI      2 2 2 2 3 4 3 4 15 10 10 10 10 2 10 0,02( ) 8 10 0,1 10 8 10 0,1 20 D rad                2 max 7,5( / )kN cm  1/10/2013 21 6.6. Example Problem 2 1/10/2013 22 6.6. Example Problem 2 1/10/2013 23 6.6. Example Problem 2 1/10/2013 24 Homework 1/10/2013 25 Homework 1/10/2013 26 Homeworks 1/10/2013 27 THANK YOU FOR ATTENTION ! 1/10/2013 28